Rotation invariant 1-forms on spheres I'm currently learning calculus on manifolds, and I find myself feeling rather uncomfortable with basic calculations to do with forms. I am therefore seeking a second opinion on my solution to the following problem:

[5.9.1, Berger and Gostiaux]
Let $\xi$ be a one-form on $\mathbb{S}^2$. Assume that $\xi$ is invariant under rotation, that is, $s^{\star} \xi = \xi$ for any $s \in \text{SO}(3)$. Prove that $\xi = 0$.

Solution:
Let $U = \mathbb{S}^2 - \{N\}$, where $N = (0,0,1)$ is the north pole, and let $f: U \to \mathbb{R}^2$ be the standard stereographic projection. Then let $\hat{\xi} := (f^{-1})^{\star} \xi$ be the coordinate representation of $\xi$ in $U$. Next, let $s \in \text{SO}(3)$ be a rotation that fixes the poles, so that it restricts to a map of $U$ into itself. Then $\hat{\xi}$ must remain invariant under the coordinate representation of $s$, since (restricting everything in the sphere to $U$)
$$(f \circ s \circ f^{-1})^{\star} \hat{\xi} = (s \circ f^{-1})^{\star} \xi = (f^{-1})^{\star} s^{\star} \xi = \hat{\xi}.$$
Now the coordinate representation of $s$ is simply the top-left $2 \times 2$ block in $s$, i.e. this is a rotation $r \in \text{SO}(2)$. Conversely, any such rotation is the coordinate representation of a rotation in $\text{SO}(3)$ (just add an extra column and row).
So let $r$ be a half-turn of the plane, and consider the effect of $(r^{\star}\hat{\xi})(0)$ on $v \in T_0 \mathbb{R}^2$. We have
$$(r^{\star}\hat{\xi})(0)(v) = \hat{\xi}(r(0))((T_0 r) v) = \hat{\xi}(0)((T_0 r)v)$$
Now since $r$ is a linear map, $T_0 r = r$, so this reduces to $-\hat{\xi}(0)(v)$. But by the $r$-invariance of $\hat{\xi}$, this is also equal to $\hat{\xi}(0)(v)$. This is true for every $v$, so $\hat{\xi}(0) = 0$. It follows (since $f$ is a diffeomorphism) that $\xi(S) = 0$.
Finally, let $p \in \mathbb{S}^2$ be arbitrary. Since $\text{SO}(3)$ acts transitively, we can find an $s$ that sends $p \leadsto S$. Then we have that $\xi(p) = (s^{\star} \xi)(p)$, and this is zero:
$$(s^{\star}\xi)(p) = (T_p s)^{\star} (\xi(s(p)) = 0,$$
and we are done.

I am moderately confident that this procedure works, but it feels like there should be cleaner solutions. Could one approach this by looking at $S^{2}$ as an embedded submanifold of $R^2$? Or is passing to charts often necessary?
Thanks in advance.
 A: At a glance this proof looks fine to me, but there's no need to pass to coordinates. Here's a coordinate-free translation of your proof that also doesn't break symmetry by fixing a point on the sphere for stereographic projection. I'll leave the second half of the argument behind a spoiler, in case you want to try to complete it yourself first.
If we fix $p \in \Bbb S^2$, consider the unique rotation $R \in O(3)$ that fixes $p$ but acts by $-\operatorname{id}$ on the plane $p^\perp \subset \Bbb R^3$. Since $2$ is even, $\det R = (-1)^2 = +1$ so in fact $R \in SO(3)$. Then, because $R$ fixes $p$, $T_p R$ is a map $T_p \Bbb S^2 \to T_p \Bbb S^2$.

As usual we can canonically identify $T_p \Bbb S^2$ with $p^\perp$, so since $R$ is linear, $T_p R$ acts by $-\operatorname{id}$ on $T_p \Bbb S^2$. In particular, $\xi_p = R^* \xi_p = -\xi_p$, so $\xi_p = 0$. Since $p$ is arbitrary $\xi = 0$.

Remark This proof works essentially unchanged for $\Bbb S^n$ for any even $n \geq 2$. For odd $n$, $R$ will have $\det R = (-1)^n = -1$, so we only have $R \in O(3)$, not $R \in SO(3)$, but we can modify our proof slightly to show that the claim is still true for odd $n$.
Remark Here's an alternative proof that only works for even $n$:
Raising indices with the standard Riemannian metric $g$ on $\Bbb S^n$, we get a invariant vector field $\xi^\sharp$ on $\Bbb S^n$ characterized by $\xi_p(Y) = g(\xi^\sharp_p, Y)$ for all $p$. Now, the Hairy Ball Theorem implies that $\xi^\sharp_p = 0_p$ for some $p$. Since $SO(n + 1)$ acts transitively on $\Bbb S^n$, invariance gives $\xi^\sharp_q = 0_q$ for all $q$, so $\xi^\sharp = 0$ and hence $\xi = 0$.
